\(A=x^2+6x+5=\left(x^2+6x+9\right)-4=\left(x+3\right)^2-4\ge-4\)
Vậy \(MIN_A=-4\) khi \(\left(x+3\right)^2=0\Leftrightarrow x=-3\)
\(B=\left(x-1\right)\left(x-3\right)=x^2-4x+3=\left(x^2-4x+4\right)-1=\left(x-2\right)^2-1\ge-1\)
Vậy \(MIN_B=-1\) khi \(\left(x-2\right)^2=0\Leftrightarrow x=2\)
\(C=x^2-x+8=\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{31}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{31}{4}\ge\dfrac{31}{4}\)
Vậy \(MIN_C=\dfrac{31}{4}\) khi \(\left(x-\dfrac{1}{2}\right)^2=0\Leftrightarrow x=\dfrac{1}{2}\)
\(D=x^2-3x=\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{4}=\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{4}\ge-\dfrac{9}{4}\)
Vậy \(MIN_D=-\dfrac{9}{4}\) khi \(\left(x-\dfrac{1}{2}\right)^2=0\Leftrightarrow x=\dfrac{1}{2}\)
